Suppose $L$ has a regular parametrix . Assume $U$ is a distribution given in an open set $\Omega \subset R^d$ and $L(U)=f$ , with $f$ a $C^{\infty}$ function in $\Omega$ , then $U$ agrees with a $C^{\infty}$ function on $\Omega$

I have learned this theorem in Stein's functional analysis Page$_{133}$ and the author proved this theorem in the following step .

$(1)$ Show that it suffices to prove $U$ agrees with a $C^\infty$ function on any ball $B$ with its closure $B^* \subset \Omega$ .

$(2)$ Show that for any ball $B$ with its closure $B^* \subset \Omega$ , we can find a function $f_B \in C^{\infty}$ which depends on the ball $B$ such that $U$ agree with $f_B$ on $B$ .

After finished the above two step , the author state that the theorem is proved , but I did not see it .

My attempt :
For step $(1)$ , assume $U$ argees with $g$ . We need to prove for all $\varphi \in D(\Omega)$ , we have $$U(\varphi)=\int g(x)\varphi(x) \,dx$$ If this is not true for some $\varphi$ . We can find a function $\phi$ such that $L\phi=\varphi$ . Then $$\int g(x)\varphi(x) \,dx \neq U(\varphi)=L(U)(\phi)=\int f(x) \phi(x) \,dx$$ WLOG , we have $g(x)\varphi(x) \lt f(x) \phi(x)$ in some open ball $O$ . Then $$\int g(x) \varphi(x) \chi_O(x) \,dx \lt \int f(x)\phi(x) \chi_O(x) \,dx$$
Notice that $\varphi(x)\chi_O(x)=L(U)(\phi(x)\chi_O(x))$ . However , these two functions are not in $C^\infty$ . So I want to show $$U(\varphi(x)\chi_O(x))=\int g(x) \varphi(x)\chi_O(x) \,dx$$ and $$L(U)(\varphi(x)\chi_O(x))=\int f(x) \varphi(x)\chi_O(x) \,dx$$

My question :
$(a)$ Can we show step $(1)$ follow the discussion above ? If we do not have the assumption $L(U)=f$ , does $(1)$ still valid ?

$(b)$ In part $(2)$ , for each ball $B$ , we have different functions $f_B \in C^{\infty}(\Omega)$ , so how to use $(2)$ to show $U$ agrees with a function $f$ ?


For each ball $B_{\alpha} $ , let $f_{\alpha} $ denote the function which agrees with distribution $U$ on $B_{\alpha}$ . Since $\Omega$ is open , it can be write as a union (might not disjoint) of open balls $\bigcup B_{\alpha}$ . We now define the function $f$ as follows .
For each $x$ , we can find a ball $B_{\alpha}$ contains $x$ ,and we define $f(x)=f_{\alpha}(x)$ . To see $f$ is well defined , WLOG, assume $B_{\beta}$ is another ball contains $x$ and $f_{\beta}(x)\lt f_{\alpha}(x)$ . Then we can find a ball $C$ contains $x$ with $C \subset B_{\alpha}$ and $C\subset B_{\beta}$ such that whenever $t \in C$ we have $f_{\beta}(t)\lt f_{\alpha}(t)$ . Find a positive function $\phi \in D(\Omega)$ with $\phi(x)\gt 0$ and supported in $C$ , then we have $$U(\phi)=\int f_{\alpha}(t)\phi(t) \, dt=\int f_{\beta}(t)\phi(t) \, dt$$ which is impossible . So the function $f$ is well defined . Also , note that for each $x\in \Omega$ , we can find a ball $B_{\alpha}$ contains $x$ , and $f(x)=f_{\alpha}(x)$ whenever $x \in B_{\alpha}$ . So $f \in C^{\infty}(\Omega)$ .

To see $$F(\varphi)=\int f(x)\varphi(x)\,dx$$ whenever $\varphi \in D(\Omega)$ . Since the support $K$ of $\varphi$ is compact , we can find finite union of balls to cover $K$ , we say $K\subset \bigcup_{n=1}^N B_n$ . Then by the partition of unity , we can find function $\eta_k$ $(1\le k\le N)$ $\in D$ with $supp(\eta_k)\subset B_k$ and $\sum_{n=1}^N\eta_n(x)=1$ whenever $x\in K$ . Then we have $$F(\varphi)=F(\sum_{n=1}^N \varphi(x)\eta_n(x))=\sum_{n=1}^N F( \varphi(x)\eta_n(x))=\sum_{n=1}^N \int \varphi(x) \eta_n(x) f(x) \,dx \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, =\int f(x)\varphi(x) \,dx$$ And the proof is completed .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.